If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

Main content

Systems of linear equations word problems — Harder example

Watch Sal work through a harder Systems of linear equations word problem.

Want to join the conversation?

  • blobby green style avatar for user najatmasri5
    you can solve it with less time and effort. you can substract 4 children and 1 adult from each choice and see if the remaining children are double the adults since each of the remaining adults brought 2 children. think smarter not harder haha
    (52 votes)
    Default Khan Academy avatar avatar for user
  • aqualine ultimate style avatar for user Jashan Nagra
    Agnes has 23 collectible stones, all of which are labradorite crystals or galena crystals. Labradorite crystals are worth $20 each, while galena crystals are worth $13 each. Agnes earns $439 by selling her entire collection. How many stones of each type did she sell?


    I'm still stuck on this one problem....please help!
    (19 votes)
    Default Khan Academy avatar avatar for user
  • female robot amelia style avatar for user amna amna
    good concept but a little bit confusing
    (29 votes)
    Default Khan Academy avatar avatar for user
  • male robot hal style avatar for user Parsa
    That was a tricky one!
    (24 votes)
    Default Khan Academy avatar avatar for user
  • duskpin sapling style avatar for user Paulina B.
    Why are we given harder practice problems then what the example shows? This one looks more simple, but the other practice problems are so complicated sometimes. The SAT is coming soon, I just feel so under prepared..
    (17 votes)
    Default Khan Academy avatar avatar for user
  • blobby green style avatar for user Aryan  Gupta
    My solution:

    Consider the number of adults to be x
    Subsequently, the number of children can be written in terms of x as "2x + 2". (Each adult brings 2 children and one of the adults brought 2 extra children)

    Now multiply the number of adults and children with their corresponding prices and equate that to 60.
    That is "2(2x + 2) + 4x = 60"

    Upon solving, you shall get x = 7 which implies that there are 7 adults. Thereafter, you can also find out the number of children which is "2x + 2" and that is 16. Thus, Option C is correct.
    (14 votes)
    Default Khan Academy avatar avatar for user
  • blobby green style avatar for user tjohnson52717
    A lil confusing
    (8 votes)
    Default Khan Academy avatar avatar for user
  • blobby green style avatar for user nathancolexd
    Here's my solution:

    4+2(a-1)=c
    4+2a-2=c
    2a+2=c

    2(2a+2)+4a=60
    4a+4+4a=60
    8a=56
    a=7

    *Note: a-1 = number of children who brought 2 children
    subtract the one who brought 4
    (8 votes)
    Default Khan Academy avatar avatar for user
  • duskpin seedling style avatar for user Charlotte Dyball
    how can we use this for graphs?
    (6 votes)
    Default Khan Academy avatar avatar for user
  • blobby green style avatar for user danielzutto
    Hi, is this problem you immediately knew that there will be only 1 adult with two kids so you did the math of 4+2x2=8.
    I thought, however, that there will be two adults (parents) as my fist logical thing to think.
    how could one understand that there is only one adult for each 2 kids?

    thanks!
    (4 votes)
    Default Khan Academy avatar avatar for user

Video transcript

- [Instructor] Tickets for a play were $2 for each child and $4 for each adult. At one showing of the play, one adult brought four children and the remaining adults brought two children each. The total ticket sales from the children and adults was $60. How many children and adults attended the play? Alright, this is an interesting one. Okay, so let's just think about how much we've spent at the play and we know it has to add up to $60. Let's think about it in terms of the children and the adults and their admissions. So you have this one adult right over here that brought four children. So how much is that adult, how much is this family? Let's just assume it's a family. How much are they going to spend? Well, that one adult is going to spend $4 for their own ticket and then four children at $2 each. So plus four children times $2 per child, this is going to be $8 for the children's tickets plus $4 on theirs. They're going to spend $12. So that adult is going to spend $12. And then there's some remaining number of adults that brought two children. So let's just say r is I could say the remaining number of adults or the number of adults with two children. Adults with two children, that's r. So each of these adults with two children, how much are they going to spend? Well, they're each going to spend $4 on their own ticket for the adult and then they're gonna have two children at $2 each, so they're gonna spend $4 on the children's tickets. So they're gonna spend $8 in total. So each of these adults with two children is gonna spend $8 at the play and there's r of them. So they're going to spend $8 for each of these adults with two children and there are r of them. So this is the total amount of ticket sales from the adults with two children. And we add that to the ticket sales from this one adult with the four children and they're gonna have to add up to $60. So this is gonna have to add up to $60. Let's see, we can subtract 12 from both sides and so on the left, we'll be left with 8r is equal to 60 minus 12 is 48. Divide both sides by eight and you get r is equal to six. So we wanna be very careful, you might say, okay there was six adults these are just the adults with two children. There's six adults with two children, but there's another adult. There's another adult who brought four children. So there's a total of seven adults, seven adults total. Pardon my handwriting. Seven adults total. So we could just look at these choices, only one of these choices have seven adults. And we could verify that this would also amount to 16 children because this person up here, in magenta, they would bring four children, so you would have four children plus six adults brought two children, so six adults bringing two children each, that would amount to 12 children. And that indeed, that indeed does add up to be 16. And if you're under time pressure, you can see there is only one choice that has seven adults, so you could just pick that one.